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3 years ago

How much energy is required to boil 65 grams of 100°C water And then heat the steam to 150°C?

Chemistry

1 answer:

Artyom0805 [142]3 years ago

8 0

Answer:

13598 J

Explanation:

Q = m × c × ∆T

Where;

Q = amount of energy (J)

m = mass (grams)

c = specific heat capacity

∆T = change in temperature

m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C

Q = 65 × 4.184 × (150 - 100)

Q = 271.96 × 50

Q = 13598 J

Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.

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kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

3 years ago

assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.

dimaraw [331]

Answer:

V = 22.41 L

Explanation:

Given data:

Mass of nitrogen = 14.0 g

Volume of gas at STP = ?

Gas constant = 0.0821 atm.L/mol.K

Solution:

Number of moles of gas:

Number of moles = mass/molar mass

Number of moles= 14 g/ 14 g/mol

Number of moles = 1 mol

Volume of gas:

PV = nRT

1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K

V = 22.41 atm.L / 1 atm

V = 22.41 L

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The latent heat of melting is 80 calories for 1 gram of ice. if 1 gram of ice absorbs 60 calories what do i get?

Eva8 [605]

Latent heat of melting is the energy that a solid absorbs to change its phase as its liquid. During this process, since all energy is used to change the phase, the temperature is constant.

Here the latent energy of melting for 1 g of ice is 80 calories and that 1 g of ice only absorbed 60 calories. hence the phase is not changed because it requires more 20 calories to melt.

Hence 1 g of ice remains as its solid phase (ice).

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3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

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Kisachek [45]

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